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Covariant/contravariant transform and metric tensor

  1. Sep 7, 2012 #1
    H is a contravariant transformation matrix, M is a covariant transformation matrix, G is the metric tensor and G-1 is its inverse. Consider an oblique coordinates system with angle between the axes = α

    I have G = 1/sin2α{(1 -cosα),(-cosα 1)} <- 2 x 2 matrix

    I compute H = G*M where M = {(1 0), (cosα sinα)} and get H = {(1 -1/tanα),(0 1/sina)} which is what I expect.

    Now I want go from H back to M so I compute M = G-1H

    So by my reckoning G-1 = 1/sin4α{(1 cosα),(cosα 1)}

    But when I multiply G-1 and H I don't get back to M. The is a 1/sin4α multiplying the whole thing.
    What am I missing?
     
  2. jcsd
  3. Sep 7, 2012 #2
    When you say G is the metric tensor, the metric tensor converts contravariant to covariant? Or is it the other way around?

    Actually, I think it's pretty clear that the issue is with your computed [itex]G^{-1}[/itex]. The operator [itex]G[/itex] has determinant 1. You've got an extra factor of [itex]1/\sin^2 \alpha[/itex] for some reason. Did you forget the prefactor from [itex]G[/itex] somewhere?
     
    Last edited: Sep 7, 2012
  4. Sep 8, 2012 #3
    contravariant = G * covariant
    covariant = G-1 * contravariant

    If G = 1/sin2α{(1 -cosα),(-cosα 1)}

    then wouldn't G-1 = (1/sin2α)(1/|detG|){(1 +cosα),(+cosα 1)}

    which gives 1/sin4α{(1 +cosα),(+cosα 1)}

    Now if I compute GG-1 I get 1/sin4{(sin2α 0),(0 sin2α)} which is equivalent to 1/sin2α{(1 0),(0 1)}

    So I get the identity matrix multiplied by 1/sin2α which doesn't add up.

    Incidentally, this question came up as a result of watching this video http://www.youtube.com/watch?v=qDLmJE2bOy4
     
  5. Sep 8, 2012 #4
    You're saying [itex]\det G = 1/\sin^2 \alpha[/itex]? Because it's not.
     
  6. Sep 8, 2012 #5
    No, I thought I had 1/|detG| where |detG| = 1 - cos2α = sin2α
     
  7. Sep 8, 2012 #6
    It's not [itex]\sin^2 \alpha[/itex] either. Why don't you go through the whole calculation of the determinant?
     
  8. Sep 8, 2012 #7
    G = 1/sin2a{(1 -cosa),(-cosa 1)} = {(1/sin2a -cosa/sin2a),(-cosa/sin2a 1/sin2a)}
    detG = 1/sin4a - cos2a/sin4a = 1/sin2a
    Thus, 1/|detG| = sin2a
    So G-1 = sin2a{(1 +cosa),(+cosa 1)}
    Now GG-1 does equal the identity matrix. Am I good so far? If so I will return to my original question in the next post.
     
  9. Sep 8, 2012 #8
    Okay, now that you're getting the identity for [itex]GG^{-1}[/itex], I expect the problems you were having will be resolved.
     
  10. Sep 9, 2012 #9
    Yes, I got it to work when I used the form G-1 = sin2α(1 +cosα),(+cosα 1)}. If I use G-1 = {(sin2α +cosα.sin2α),(cosα.sin2α,1)} it doesn't.
     
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